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bai2 :cmr
a, a^3+b^3=(a+b)^3-3ab.(a+b)
VP= \(\left(a+b\right)^3-3ab\left(a+b\right)\)
=\(a^3+b^3+3a^2b+3ab^2-3a^2b-3ab^2=a^3+b^3\)
=VT
b.a^3-b^3=(a-b)^3+3ab,(a-b)
\(VP=\left(a-b\right)^3+3ab\left(a-b\right)\)
=\(a^3-3a^2b+ab^2.3-b^3+3a^2b-3ab^2=a^3-b^3\)
=VT
=> ĐPCM
bài 1.
a) = 8x^3+4x^2y+2xy^2-4x^2y-2xy^2-y^3-(8x^3-4x^2y+2xy^2+4x^2y-2xy^2+y^3)
= 8x3+4x2y+2xy2-4x2y-2xy2-y3 - 8x3+4x2y-2xy2-4x2y+2xy2-y3
=-8x2y-6y3
b) = 27x3-18x2y+12xy2+18x2y-12xy2+8y3-27x3
=8y
a) \(\dfrac{2x-2y}{x^2-2xy+y^2}=\dfrac{2\left(x-y\right)}{\left(x-y\right)^2}=\dfrac{2}{x-y}\)
b) \(\dfrac{2-2a}{a^3-1}=-\dfrac{2-2a}{1-a^3}=-\dfrac{2\left(1-a\right)}{\left(1-a\right)\left(1+a+a^2\right)}=\dfrac{-2}{\left(1+a+a^2\right)}\)
c) \(\dfrac{x^2-6x+9}{x^2-8x+15}\)
\(=\dfrac{x^2-2.x.3+3^2}{x^2-3x-5x+15}\)
\(=\dfrac{\left(x-3\right)^2}{\left(x^2-3x\right)-\left(5x-15\right)}\)
\(=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)-5\left(x-3\right)}\)
\(=\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x-5\right)}\)
\(=\dfrac{x-3}{x-5}\)
d) \(\dfrac{x^4-2x^3}{2x^4-x^3}\)
\(=\dfrac{x^3\left(x-2\right)}{x^3\left(2x-1\right)}\)
\(=\dfrac{x-2}{2x-1}\)
